Question

Two triangles have the same height. The base of one

triangle is twice that of the other. Find the difference in
their areas.​

Answer 1

`\huge\underline{\overline{\mid{\bold{\red{ANSWER}}\mid}}}`

They both have same height,

let the height of both triangles be = x

let the base one triangle = y

then the base of another trainlgle = 2y

Area of triangle = 1/2 X B X H

Area of Triangle 1

`(1)/(2) * y * x \\ = > (xy)/(2)`

Area of Triangle 2

`(1)/(2) * 2y * x \\ = > (2xy)/(2) \\ = > xy`

Difference in the area =

`xy - (xy)/(2) \\ = > (2xy - xy)/(2) \\ = > (xy)/(2)`

Hence Area of Triangle 1 is half the area of Triangle 2

Difference of 1/2 area.

Answer 2

`\huge\underline{\overline{\mid{\bold{\red{ANSWER}}\mid}}}∣ANSWER∣</p><p></p><p>They both have same height,</p><p></p><p>let the height of both triangles be = x</p><p></p><p>let the base one triangle = y</p><p></p><p>then the base of another trainlgle = 2y</p><p></p><p>Area of triangle = 1/2 X B X H</p><p></p><p>Area of Triangle 1</p><p></p><p>\begin{gathered} (1)/(2) * y * x \\ = > (xy)/(2) \end{gathered}21×y×x=>2xy</p><p></p><p>Area of Triangle 2</p><p></p><p>\begin{gathered} (1)/(2) * 2y * x \\ = > (2xy)/(2) \\ = > xy\end{gathered}21×2y×x=>22xy=>xy</p><p></p><p>Difference in the area =</p><p></p><p>\begin{gathered}xy - (xy)/(2) \\ = > (2xy - xy)/(2) \\ = > (xy)/(2) \end{gathered}xy−2xy=>22xy−xy=>2xy</p><p></p><p>Hence Area of Triangle 1 is half the area of Triangle 2</p><p></p><p>Difference of 1/2 area.</p><p></p><p>`

Explanation:

side it

siso,;-):-(